Help with Simplifying leaving in index form

Skye8886

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Joined
May 17, 2020
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4
Hello, I am Skye.
I am 8 and my Mum is always giving me extension work and getting me to do maths from higher grades.
Well, I think I understand the Simplifying leaving in index form process but I am not quite sure.
I have a question that I have had trouble more than most that I was hoping someone would be able to make sure I am doing it right and have got the right answer.
Here is the problem –

54 x 56
________
(53)3

This is what I got with working –
4 x 6 = 24
3 x 3= 9

24-9 = 15
= 515
Can someone tell me if this is right? And if it isn't what the answer is with working?
 

Romsek

Full Member
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Nov 16, 2013
Messages
962
Hi Skye. It's not quite correct.

\(\displaystyle 5^4 \times 5^6 = 5^{4+6} = 5^{10}\)

\(\displaystyle \left(5^3\right)^3 = 5^{3\times 3} = 5^9\)

\(\displaystyle \dfrac{5^{10}}{5^9} = 5^{10-9}=5^1 = 5\)
 

Jomo

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Dec 30, 2014
Messages
7,035
You can check your answer by using the definitions for exponents.

\(\displaystyle \dfrac{5^4 5^6}{(5^3)^3}= \dfrac{(5*5*5*5) *(5*5*5*5*5*5)}{(5*5*5)^3} = \dfrac{5*5*5*5 *5*5*5*5*5*5}{5*5*5 * 5*5*5 * 5*5*5} = 5\)
 
Last edited:

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
7,558
Hello, I am Skye.
I am 8 and my Mum is always giving me extension work and getting me to do maths from higher grades.
Well, I think I understand the Simplifying leaving in index form process but I am not quite sure.
I have a question that I have had trouble more than most that I was hoping someone would be able to make sure I am doing it right and have got the right answer.
Here is the problem –

54 x 56
________
(53)3

This is what I got with working –
4 x 6 = 24
3 x 3= 9

24-9 = 15
= 515
Can someone tell me if this is right? And if it isn't what the answer is with working?
A good way to think about this sort of thing when you are first learning (especially at your age) is to think of \(\displaystyle 5^4\) as meaning \(\displaystyle 5\times 5\times 5\times 5\). So the numerator of your example is the product of 4 \(\displaystyle 5\)'s, times another 6 \(\displaystyle 5\)'s, for a total of 10 \(\displaystyle 5\)'s. Do you see that? This is why you add exponents.

Then do something similar to the denominator.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
4,650
You are 8, which is a young age to learn about abstraction and generalization.

Exponents started out as a way to abbreviate expressions like 7 * 7 * 7 * 7 * 7. That takes time to write, and it is easy to forget one of the 7's and screw things up. So the exponent was born.

\(\displaystyle 7^5 = 7 * 7 * 7 * 7 * 7.\)

It started just as a way to make notation briefer and less prone to error. Make sense?

However that means

\(\displaystyle 7^3 * 7^4 = (7 * 7 * 7) * (7 * 7 * 7 * 7) = 7 * 7 * 7 * 7 * 7 * 7 * 7 = 7^7 = 7^{3+4}.\)

Then people made a discovery. It turns out that it doesn't make any difference whether the base number is 7 or 11 or 1/2, and it does not make any difference whether the exponents are 3 and 4 or 7 and 11. It is universally true that if a, b, and c are ANY numbers greater than zero and exponents mean what we have said they mean then

\(\displaystyle a^b * a^c = a^{b+c}.\)

If you want to test this FIRST LAW OF EXPONENTS on your calculator, go ahead.

\(\displaystyle 2^3 = 2 * 2 * 2 = 8.\\
2^5 =2 * 2 * 2 * 2 *2 = 4 * 4 * 2 = 16 * 2 = 32.\\
\text {So } 2^3 * 2^5 = 8 * 32 = 256. \implies 2^3 * 2^5 = 8 * 32 = 256.\\
2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 4 * 2 * 2 * 2 * 2 * 2 * 2 = 8 * 2 * 2 * 2 * 2 * 2 =\\
16 * 2 * 2 * 2 * 2 = 32 * 2 * 2 * 2 = 64 * 2 * 2 = 128 * 2 = 256.\)

So exponents are a handy way to save time and avoid errors by reducing work. Then someone figured out that


\(\displaystyle (2^3)^4 = (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) * (2 * 2 * 2) =\\

2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 2^{12} = 2^{3 * 4}.\)
Again it turns out that the principle is true no matter what the base or exponents are. We then get the

SECOND LAW OF EXPONENTS, namely, if a, b, and c are ANY numbers greater than zero,

\(\displaystyle (a^b)^c = a^{b*c}.\)

Did you follow all this?

If so, I should tell you that with a major adjustment to the definition of exponent, we get to keep the meanings you already know and the two laws shown above, but we get some additional neat ways to simplify our work and notation. So the modern mathematician does not define exponents the way you have learned. The modern definition gives the same results as your definition, but gives extra results as well.
 

Skye8886

New member
Joined
May 17, 2020
Messages
4
Hi Skye. It's not quite correct.

\(\displaystyle 5^4 \times 5^6 = 5^{4+6} = 5^{10}\)

\(\displaystyle \left(5^3\right)^3 = 5^{3\times 3} = 5^9\)

\(\displaystyle \dfrac{5^{10}}{5^9} = 5^{10-9}=5^1 = 5\)

Thank you so much, I understand where I went wrong now. I forgot the index laws and multiplied instead of adding.
:)
 

Skye8886

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Joined
May 17, 2020
Messages
4
A good way to think about this sort of thing when you are first learning (especially at your age) is to think of \(\displaystyle 5^4\) as meaning \(\displaystyle 5\times 5\times 5\times 5\). So the numerator of your example is the product of 4 \(\displaystyle 5\)'s, times another 6 \(\displaystyle 5\)'s, for a total of 10 \(\displaystyle 5\)'s. Do you see that? This is why you add exponents.

Then do something similar to the denominator.
Yes, thank you I see what you are explaining. Thank you so much for this it is really helpful.
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
2,397
… index form …
… index laws
Hi Skye8886. I'm not sure where you're posting from, but the terminology in the USA differs from 'index'.

Just so you know, we say 'exponential form' and 'exponent laws'. The word 'index' is used for other things, in our math vocabulary.

😎
 

Skye8886

New member
Joined
May 17, 2020
Messages
4
H
Hi Skye8886. I'm not sure where you're posting from, but the terminology in the USA differs from 'index'.

Just so you know, we say 'exponential form' and 'exponent laws'. The word 'index' is used for other things, in our math vocabulary.

😎
Hi,

I am in Australia that is why I call it something different. I respect that your terminology is different, this is just how I have been taught.

Skye.
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
2,397
… [my] terminology is different, this is just how I have been taught …
Hi Skye. I understood that. I'm just giving you a 'heads-up', so that you'll be aware going forward that your terminology might confuse people (located in places where 'index' has a different meaning than 'exponent'). Cheers!

\(\;\)
 

Harry_the_cat

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Joined
Mar 16, 2016
Messages
1,951
Thank you so much, I understand where I went wrong now. I forgot the index laws and multiplied instead of adding.
:)
Skye, you sound like a very smart young student. Just a hint when learning maths. Don't aim to "remember" things, but aim to "understand" them and know where they come from. You will never "forget" that index law if you understand WHY you add them in that case.
Good luck with your studies. PS I'm an Aussie too.
 

Otis

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Apr 22, 2015
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\(\displaystyle \sqrt[n]{x}\)

Now, I'm curious. That number n is an example of what we call an index (the index of the radical.) Different name in Straya?

😎
 

Harry_the_cat

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Messages
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We use the word index or exponent interchangeably when dealing with powers. Of course, index has another meaning as in Price index, seasonal index, etc.
What you showed Otis, we call the nth root of x, don't you? No specific name for the n that I know of. It could of course be written as x^ (1/n) and then 1/n would be the index (or exponent).
 

Otis

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Messages
2,397
…What you showed Otis, we call the nth root of x, don't you? … 1/n would be the index [in the expression x1/n] …
Interesting. Our radical index is the reciprocal of your exponential index, which is our exponent (the reciprocal of our radical index).

There's probably a name in there somewhere, for down under. Whatever it may be, Aussies will shorten it. So, how about 'radix'?

;)
 
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